# Chapter 14: ALTERNATORS AND SYNCHRONOUS MOTORS¶

## Example 14.1,Page number: 433¶

In [1]:
#Question:
"""Finding the number of poles on the generator if the frequency of the generated voltage is decreased."""

#Variable Declaration:
f=60                  #Frequency of ac-generator(in Hertz)
P=6                   #Number of poles

#Calculations:
Ns=(120*f)/P
f=20
P=(120*f)/Ns

#Result:
print "The speed of rotation of the generator is %d rpm." %(Ns)
print "If the frequency is reduced to 20Hz,the required number of poles is %d." %(P)

The speed of rotation of the generator is 1200 rpm.
If the frequency is reduced to 20Hz,the required number of poles is 2.


## Example 14.2,Page number: 441¶

In [5]:
#Question:
"""Finding the distribution factor for a machine."""

from math import radians,degrees,sin

#Variable Declaration:
slots=9                      #Number of slots

#Calculations:
q_a=120.0/degrees(slot_angle)
k_d_a=sin(q_a*(slot_angle/2.0))/(q_a*sin(slot_angle/2.0))
q_b=60.0/degrees(slot_angle)
k_d_b=sin(q_b*(slot_angle/2.0))/(q_b*sin(slot_angle/2.0))

#Result:
print "(a)The distribution factor for a machine for a three phase winding with 120 degrees phase group is %.3f." %(k_d_a)
print "(b)The distribution factor for a machine for a three phase winding with 60 degrees phase group is %.3f." %(k_d_b)

(a)The distribution factor for a machine for a three phase winding with 120 degrees phase group is 0.831.
(b)The distribution factor for a machine for a three phase winding with 60 degrees phase group is 0.960.


## Example 14.3,Page number: 441¶

In [6]:
#Question:
"""Finding the speed,the generated emf per phase,and the line emf of a three-phase alternator."""

from math import radians,degrees,sqrt,sin

#Variable Declaration:
phase=3                #Number of phases
f=50                   #Frequency rating(in Hertz)
P=20                   #Number of poles
slots=180              #Number of slots on the stator
cond_per_slot=8        #Number of conductors per slot
flux=25e-03            #Flux per pole(in Weber)

#Calculations:
Z=slots*cond_per_slot
T=(Z/2)/phase
Ns=(120*f)/P
k_p=1
slots_per_pole=slots/P
q=slots_per_pole/phase
k_d=sin(q*(slot_angle/2))/(q*sin(slot_angle/2))
E=4.44*f*flux*T*k_p*k_d
line_emf=sqrt(3.0)*E

#Result:
print "(a)The speed of the alternator is %d rpm." %(round(Ns,0))
print "(b)The rms value of generated EMF per phase is %.2f V." %(E)
print "(c)The line EMF is %.2f V." %(line_emf)

(a)The speed of the alternator is 300 rpm.
(b)The rms value of generated EMF per phase is 1278.45 V.
(c)The line EMF is 2214.34 V.


## Example 14.4,Page number: 446¶

In [14]:
#Question:
"""Finding the voltage regulation for full-load for a three-phase alternator."""

from math import sqrt,cos,acos

#Variable Declaration:
P=600e06              #Power rating of the alternator(in VA)
V_L=22e03             #Rated terminal voltage(in Volts)
sync_imp=0.16         #Synchronous impedance per phase(in Ohms)
res_phase=0.014       #Resistance per phase(in Ohms)

#Calculations:
I_L=P/(sqrt(3)*V_L)
Iph=I_L
V=V_L/sqrt(3)
Vz=Iph*sync_imp
theta=acos(res_phase/sync_imp)
pf=0.8
phi=acos(pf)
alpha=theta-phi
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
vol_reg_a=(E-V)/V
pf=1
phi=acos(pf)
alpha=theta-phi
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
vol_reg_b=(E-V)/V
pf=0.8
phi=acos(pf)
alpha=theta+phi
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
vol_reg_c=(E-V)/V

#Result:
print "(a)For a power factor of 0.8 lagging:"
print "The voltage regulation for full load is %.2f percent." %(vol_reg_a*100)
print "(b)For unity power factor:"
print "The voltage regulation for full load is %.2f percent." %(vol_reg_b*100)
print "(c)For a power factor of 0.8 leading:"
print "The voltage regulation for full load is %.2f percent." %(vol_reg_c*100)
print "\nNote:The voltage regulation for leading power-factor load is negative."
print "It means that on removing the load, the terminal voltage decreases."

(a)For a power factor of 0.8 lagging:
The voltage regulation for full load is 14.20 percent.
(b)For unity power factor:
The voltage regulation for full load is 3.64 percent.
(c)For a power factor of 0.8 leading:
The voltage regulation for full load is -8.90 percent.

Note:The voltage regulation for leading power-factor load is negative.
It means that on removing the load, the terminal voltage decreases.


## Example 14.5,Page number: 450¶

In [16]:
#Question:
"""Finding the synchronous reactance per phase and the voltage regulation for a three-phase star-connected alternator."""

from math import acos,sqrt,cos

#Variable Declaration:
V_L=900               #Open-circuit voltage(line to line)(in Volts)
V_L_rated=3.3e03      #Rated voltage of the alternator(in Volts)
I_f=100               #Full-load current(in Amperes)
R=0.9                 #Armature Resistance(in Ohm/phase)

#Calculations:
V_oc=V_L/sqrt(3)
I_sc=I_f
sync_imp=V_oc/I_sc
sync_rea=sqrt((sync_imp*sync_imp)-(R*R))
theta=acos(R/sync_imp)
V=V_L_rated/sqrt(3)
Vz=I_f*sync_imp
pf=0.8
phi=acos(pf)
alpha=theta-phi
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
vol_reg_a=(E-V)/V
pf=0.8
phi=acos(pf)
alpha=theta+phi
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
vol_reg_b=(E-V)/V

#Result:
print "The synchronous reactance per phase is %.3f Ohms." %(sync_rea)
print "(a) For a power factor of 0.8 lagging:"
print "The voltage regulation for full load is %.2f percent" %(vol_reg_a*100)
print "(b) For a power factor of 0.8 leading:"
print "The voltage regulation for full load is %.2f percent." %(vol_reg_b*100)
print "\nNote: The voltage regulation for leading power-factor load is negative."
print "It means that on removing the load, the terminal voltage decreases."


The synchronous reactance per phase is 5.118 Ohms.
(a) For a power factor of 0.8 lagging:
The voltage regulation for full load is 21.34 percent
(b) For a power factor of 0.8 leading:
The voltage regulation for full load is -9.03 percent.

Note: The voltage regulation for leading power-factor load is negative.
It means that on removing the load, the terminal voltage decreases.


## Example 14.6,Page number: 455¶

In [17]:
#Question:
"""Finding the angle of retard of a synchronous motor."""

from math import sqrt,acos,cos,pi,asin,sin,degrees
from cmath import phase

#Variable Declaration:
Po=9e03               #Power rating of synchronous motor(in Watts)
V_L=400               #Voltage rating of synchronous motor(in Watts)
Zs=0.4+3*1j           #Synchronous impedance per phase(in Ohms)
effi=0.9              #Efficiency of the motor

#Calculations:
Pin=Po/effi
I_L=Pin/(sqrt(3)*V_L*pf)
I=I_L
phi=acos(pf)
mod_Zs=abs(Zs)
theta=phase(Zs)
V=V_L/sqrt(3)
Er=I*mod_Zs
E=sqrt((V*V)+(Er*Er)+(2*V*Er*cos(pi-(theta+phi))))
E_L=sqrt(3)*E
angle_retard=asin((Er*sin(theta+phi))/E)

#Result:
print "The angle of retard of the rotor is %.2f degrees." %(degrees(angle_retard))
print "The excitation emf E to which the motor has to be excited to give a full-load output at 0.8 leading power factor is %.2f V." %(E_L)

The angle of retard of the rotor is 10.47 degrees.
The excitation emf E to which the motor has to be excited to give a full-load output at 0.8 leading power factor is 453.81 V.


## Example 14.7,Page number: 459¶

In [9]:
#Question:
"""Finding the line emf generated by the alternator."""

#Variable Declaration:
S=24.0                #Number of slots in the alternator
C=12.0                #Number of conductors per slot
flux=0.1              #Flux per pole(in Weber)
P=4.0                 #Number of poles in the alternator
Ns=1500.0             #Synchronous speed of the alternator(in rpm)

#Calculations:
Zph=S*C/3.0
T=Zph/2.0
S_pole=S/P
slot_ang=180.0/S_pole
q=S_pole/3.0
f=(P*Ns)/120.0
kp=1.0
E=4.44*flux*f*T*kp*kd
E_L=sqrt(3.0)*E

#Result:
print "The line emf generated when the alternator runs at 1500 rpm is %.2f V." %(E_L)

The line emf generated when the alternator runs at 1500 rpm is 1782.78 V.


## Example 14.8,Page number: 460¶

In [20]:
#Question:
"""Finding the net emf induced in the 6 coil in series constituting the alternator winding."""

from math import radians,sin

#Variable Declaration:
q=6.0                 #Number of slots per pole per phase
angle=30.0            #Electrical angle between two consecutive slotss(in degrees)
e=10.0                #Emf of each coil(in Volts)

#Calculations:
arith_sum=6*e
Er=kd*arith_sum

#Result:
print "The net emf induced in the six coils in series is %.3f V." %(round(Er,3))

The net emf induced in the six coils in series is 38.637 V.


## Example 14.9,Page number: 460¶

In [21]:
#Question:
"""Finding the number of poles and the current rating of an alternator."""

from math import sqrt,pi

#Variable Declaration:
N=120.0               #Speed of the alternator(in rpm)
f=50.0                #Frequency of the alternator(in Hertz)
VA=100e06             #VA rating of the alternator(in Volt-Ampere)
pf=1.0                #Power factor
V_L=11e03             #Line voltage(in Volts)
effi=0.97             #Efficiency of the alternator

#Calculations:
P=(120.0*f)/N
Po=VA*pf
I_L=VA/(sqrt(3.0)*V_L)
Pin=Po/effi
tor=Pin/(2*pi*N/60.0)

#Result:
print "(a)The number of poles is %d." %(P)
print "(b)The current rating is %.2f A." %(I_L)
print "(c)The input power is %.2f MW." %(round((Pin/1000000),2))
print "(d)The prime-mover torque applied to the genrator shaft is %e Nm." %(tor)

(a)The number of poles is 50.
(b)The current rating is 5248.64 A.
(c)The input power is 103.09 MW.
(d)The prime-mover torque applied to the genrator shaft is 8.203863e+06 Nm.


## Example 14.10,Page number: 460¶

In [26]:
#Question:
"""Finding the percentage regulation for a load."""

from math import atan,radians,acos

#Variable Declaration:
V_L=11e03             #Line voltage of the star-connected alternator(in Volts)
VA=800e03             #Power rating of the alternator(in Volt-Amperes)
Rs=1.5                #Resistance per phase(in Ohms)
Xs=25.0               #Synchronous reactance(in Ohms)
pf=0.8                #Leading power factor
Po=600e03             #Output power(in Watts)

#Calculations:
Vph=V_L/sqrt(3.0)
V=Vph
Iph=Po/(sqrt(3.0)*V_L*pf)
Zs=sqrt((Rs*Rs)+(Xs*Xs))
theta=atan(Xs/Rs)
Vz=Iph*Zs
pf_ang=acos(pf)
alpha=theta+pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
reg=(E-V)/V

#Result:
print "The percentage regulation is %.3f per cent." %(reg*100)

The percentage regulation is -7.641 per cent.


## Example 14.11,Page number: 461¶

In [2]:
#Question:
"""Finding the full-load voltage regulation of the alternator."""

from math import atan,acos

#Variable Declaration:
V_L=6e03              #Line voltage of the alternator(in Volts)
VA=6000e03            #Power rating of the alternator(in Volt-Amperes)
R=0.2                 #Winding resistance per phase(in Ohms)
pf=0.8                #Lagging power factor
V_L_OC=480.0          #Line voltage in open-circuit test(in Volts)
I_f_OC=10.0           #Field current in open-circuit test(in Amperes)
I_L_SC=105.0          #Line current in short-circuit test(in Amperes)
I_f_SC=5.0            #Field current in short-circuit test(in Amperes)

#Calculations:
"""In the short-circuit test,the currents are small compared to the full-load current."""
V=V_L/sqrt(3.0)
I=VA/(3*V)
V_ph_OC=V_L_OC/sqrt(3.0)
"""Since the field current of 5 A gives an armature current of 105 A,a field current of 10 A will give an armature
current of (105*2)=210 A."""
I_ph=I_L_SC*2
Zs=V_ph_OC/I_ph
Xs=sqrt((Zs*Zs)-(R*R))
theta=atan(Xs/R)
Vz=I*Zs
pf_ang=acos(pf)
alpha=theta-pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
reg=(E-V)/V

#Result:
print "The full-load voltage regulation of the alternator at 0.8 lagging power factor is %.2f per cent." %(reg*100)

The full-load voltage regulation of the alternator at 0.8 lagging power factor is 16.73 per cent.


## Example 14.12,Page number: 462¶

In [6]:
#Question:
"""Finding the terminal voltage of the generator."""

#Variable Declaration:
f=50.0                #Frequency rating of the generator(in Hertz)
Z_p=96.0              #Number of conductors per phase
flux=0.1              #Flux per pole(in Webers)
Xs=5.0                #Synchronous reactance per phase(in Ohms)
kd=0.96               #Distribution factor for the stator winding
Z_L=10.0              #Load impedance(in Ohms)

#Calculations:
Zs=1j*Xs
kp=1.0
T=Z_p/2.0
E=4.44*f*flux*kp*kd*T
V=E/(1+(Zs/Z_L))
V_L=sqrt(3.0)*abs(V)

#Result:
print "The terminal voltage of the generator is %.2f V." %(V_L)

The terminal voltage of the generator is 1584.79 V.


## Example 14.13,Page number: 462¶

In [25]:
#Question:
"""Finding the percentage change of voltage."""

from math import atan,radians,acos

#Variable Declaration:
V_L=6.6e03            #Line voltage of the star-connected alternator(in Volts)
VA=1500e03            #Power rating of the alternator(in Volt-Amperes)
Rs=0.5                #Resistance per phase(in Ohms)
Xs=5.0                #Synchronous reactance(in Ohms)
pf=0.8                #Lagging power factor

#Calculations:
Vph=V_L/sqrt(3.0)
V=Vph
Iph=VA/(3.0*Vph)
Zs=sqrt((Rs*Rs)+(Xs*Xs))
theta=atan(Xs/Rs)
Vz=Iph*Zs
pf_ang=acos(pf)
alpha=theta-pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
reg=(E-V)/V

#Result:
print "The percentage change in voltage is %.3f per cent." %(reg*100)

The percentage change in voltage is 12.432 per cent.


## Example 14.14,Page number: 463¶

In [8]:
#Question:
"""Finding the terminal voltage and load regulation."""

from math import atan,radians,acos,asin

#Variable Declaration:
V_L=6599.0            #Line voltage of the star-connected alternator(in Volts)
Rs=0.5                #Resistance per phase(in Ohms)
Xs=5.0                #Synchronous reactance(in Ohms)
I=130.0               #Full-load current(in Amperes)

#Calculations:
E=V_L/sqrt(3.0)
Zs=Rs+1j*Xs
theta=atan(Xs/Rs)
Vz=I*abs(Zs)
pf=0.8
pf_ang=acos(pf)
alpha=theta-pf_ang
tor_ang=asin((Vz*sin(alpha))/E)
V_a=(E*cos(tor_ang))-(Vz*cos(alpha))
reg_a=(E-V_a)/V_a
pf=0.6
pf_ang=acos(pf)
alpha=theta+pf_ang
beta=pi-alpha
tor_ang=asin((Vz*sin(beta))/E)
V_b=(E*cos(tor_ang))+(Vz*cos(beta))
reg_b=(E-V_b)/V_b

#Result:
print "(a)The terminal voltage when the power factor is 0.8 lagging is %.2f V." %(V_a)
print "   The load regulation is %.2f per cent." %(reg_a*100)
print "(b)The terminal voltage when the power factor is 0.6 leading is %.2f V." %(V_b)
print "   The load regulation is %.2f per cent." %(reg_b*100)

(a)The terminal voltage when the power factor is 0.8 lagging is 3337.45 V.
The load regulation is 14.16 per cent.
(b)The terminal voltage when the power factor is 0.6 leading is 4265.21 V.
The load regulation is -10.67 per cent.


## Example 14.15,Page number: 464¶

In [4]:
#Question:
"""Finding the percentage load regulation."""

from math import acos,atan

#Variable Declration:
VA=1.5e06            #Power rating of the synchronous generator(in Volt-Amperes)
V_L=11e03            #Line voltage(in Volts)
Rs=1.2               #Armature resistance(in Ohms)
Xs=25.0              #Synchronous reactance per phase(in Ohms)
P_L=1.4375e06        #Load to be delivered(in Volt-Amperes)

#Calculations:
V=V_L/sqrt(3.0)
I=P_L/(3*V)
Zs=sqrt((Rs*Rs)+(Xs*Xs))
Vz=I*Zs
pf=0.8
pf_ang=acos(pf)
theta=atan(Xs/Rs)
alpha=theta-pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
reg_a=(E-V)/V
alpha=theta+pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
reg_b=(E-V)/V

#Result:
print "(a)The percentage load regulation for a power factor of 0.8 lagging is %.2f per cent." %(reg_a*100)
print "(b)The percentage load regulation for a power factor of 0.8 leading is %.2f per cent." %(reg_b*100)

(a)The percentage load regulation for a power factor of 0.8 lagging is 21.15 per cent.
(b)The percentage load regulation for a power factor of 0.8 leading is -13.12 per cent.


## Example 14.16,Page number: 465¶

In [5]:
#Question:
"""Finding the torque angle of the alternator."""

from math import acos,degrees

#Variable Declaration:
R=0.5                 #Effective resistance per phase of the alternator(in Ohms)
V_L=2200.0            #Line voltage of the alternator(in Volts)
I_FL=200.0            #Full-load current(in Amperes)
If=30.0               #Field current(in Amperes)
V_L_oc=1100.0         #Line-to-line voltage on open circuit(in Volts)
pf=0.8                #Lagging power factor

#Calculations:
Zs=V_L_oc/(sqrt(3.0)*I_FL)
theta=acos(R/Zs)
Vz=I_FL*Zs
V=V_L/sqrt(3.0)
pf_ang=acos(pf)
alpha=theta-pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
Po=V*I_FL*pf
tor_ang=theta-acos((Po+(V*V*cos(theta)/Zs))*(Zs/(E*V)))

#Result:
print "The torque angle of the alternator is %.2f degrees." %(degrees(tor_ang))

The torque angle of the alternator is 14.35 degrees.


## Example 14.17,Page number: 465¶

In [10]:
#Question:
""""Finding the synchronising power of the alternator."""

from math import acos

#Variable Declaration:
P=6.0                 #Number of poles
VA=3e06               #Power rating of the alternator(in Volt-Amperes)
Ns=1000.0             #Speed of operation of the alternator(in rpm)
V_L=3.3e03            #Line voltage of the load(in Volts)
pf=0.8                #Lagging power factor

#Calculations:
V=V_L/sqrt(3.0)
I=VA/(sqrt(3.0)*V_L)
IXs=0.25*V
Xs=IXs/I
Vz=I*Xs
theta=pi/2
pf_ang=acos(0.8)
alpha=theta-pf_ang
E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))
Psy=(3*E*V*sin(rotor_dis))/(Xs)
tor=(3*Psy)/(2*pi*Ns/60)

#Result:
print "(a)The synchronising power is %.2f kW." %(Psy/1000)
print "(b)The synchronising torque per mechanical degree is %.2f kNm." %(tor/1000)

(a)The synchronising power is 733.08 kW.
(b)The synchronising torque per mechanical degree is 21.00 kNm.


## Example 14.18,Page number: 466¶

In [6]:
#Question:
"""Finding the armature current and power factor."""

from math import radians,degrees
from cmath import rect,phase

#Variable Declaration:
V_L=2300.0            #Line voltage of the winding(in Volts)
f=50.0                #Operating frequency of the synchronous motor(in Hertz)
Psh=205.0             #Power delivered by the motor(in Horse Power)
ang=15.0              #Power angle(in degrees)
Xs=11.0               #Synchronous reactance(in Ohms)
effi=0.90             #Efficiency of the motor

#Calculations:
V=V_L/sqrt(3.0)
Psh=Psh*746.0
Pd=Psh/effi
pf=cos(phase(I))

#Result:
print "(a)The excitation voltage per phase is %.2f V per phase." %(E)
print "(b)The armature current is %.2f A at a phase angle of %.2f degrees." %(abs(I),degrees(phase(I)))
print "(c)The power factor is %.4f leading." %(pf)

(a)The excitation voltage per phase is 1812.83 V per phase.
(b)The armature current is 57.44 A at a phase angle of 42.05 degrees.
(c)The power factor is 0.7426 leading.


## Example 14.19,Page number: 466¶

In [1]:
#Question:
"""Finding the line current and the power factor."""

#Variable Declaration:
V_L=6600.0            #Line voltage(in Volts)
Xs=20.0               #Synchronous reactance per phase(in Ohms)
Pin=915e03            #Power consumed by the motor(in Watts)
E_L=8942.0            #Induced line emf per phase(in Volts)

#Calculations:
V=V_L/sqrt(3.0)
E=E_L/sqrt(3.0)
I_cos=Pin/(sqrt(3.0)*V_L)
BN=Xs*I_cos
NA=sqrt((E*E)-(BN*BN))
NO=NA-V
Er=sqrt((NO*NO)+(BN*BN))
I_L=Er/Xs
pf=I_cos/I_L

#Result:
print "(a)The line current is %.2f A." %(I_L)
print "(b)The power factor is %.4f leading." %(pf)

(a)The line current is 97.05 A.
(b)The power factor is 0.8247 leading.


## Example 14.20,Page number: 467¶

In [3]:
#Question:
"""Finding the power factor when the load on the motor increases."""

from math import acos,asin

#Variable Declaration:
V_L=6600.0            #Line voltage(in Volts)
Xs=15.0               #Synchronous reactance per phase(in Ohms)
Pin=500e03            #Power consumed by the motor(in Watts)
pf=0.8                #Leading power factor
Pin_new=800e03        #New Input power(in Watts)

#Calculations:
V=V_L/sqrt(3.0)
Zs=Xs
I_L=Pin/(sqrt(3.0)*V_L*pf)
Er=I_L*Zs
pf_ang=acos(pf)
alpha=(pi/2)-pf_ang
E=sqrt((V*V)+(Er*Er)+(2*V*Er*cos(alpha)))
I1_cos=Pin_new/(sqrt(3.0)*V_L)
sin_tor=(Pin_new*Xs)/(E*V)
AN=V*sin_tor
NB=E-(V*cos(asin(sin_tor)))
Er1=sqrt((AN*AN)+(NB*NB))
I1=Er1/Xs
pf=I1_cos/I1

#Result:
print "The power factor when the load on the motor increases is %.4f leading." %(pf)

The power factor when the load on the motor increases is 0.3229 leading.


## Example 14.21,Page number: 469¶

In [3]:
#Question:
"""Finding the power factor and kVA rating of the synchronous motor."""

from math import atan,acos

#Variable Declaration:
Si=500e03             #Load supplied by the three-phase system(in Volt-Amperes)
pf_i=0.5              #Lagging power factor of three-phase system
Po=100.0              #Load supplied by the synchronous motor(in Horse-Power)
pf_t=0.9              #Overall lagging power factor
effi=0.87             #Efficiency of the synchronous motor

#Calculations:
Pi=Si*pf_i
Qi=Si*sin(acos(pf_i))
Ps=(Po*746)/effi
Pt=Pi+Ps
St=Pt/pf_t
Qt=St*sin(acos(pf_t))
Qs=Qt-Qi
pf_ang=atan(Qs/Ps)
pf_s=cos(pf_ang)
Ss=sqrt((Ps*Ps)+(Qs*Qs))

#Result:
print "(a)The power faactor of the synchronous motor is %.2f leading." %(pf_s)
print "(b)The kVA rating of the synchronous motor is %.2f kVA." %(Ss/1000.0)

(a)The power faactor of the synchronous motor is 0.30 leading.
(b)The kVA rating of the synchronous motor is 283.67 kVA.